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mathforfun@yahoogroups.com, bqllpd <no_reply@...> wrote:

>

> \sum_{i=0}^n y^{n-i}*\binom{n}{i}*(a^i+b^i)=c Solve for y.

>

This is equivalent to:

(y+a)^n + (y+b)^n = c

Making various substitutions this is also equivalent to solving:

(x+1)^n + (x-1)^n = K

for x in terms of K. I don't think there's any formula except for certain values of n. I tried asking Wolfram alpha to solve

(x+1)^5 + (x-1)^5 = K

for x and there was no formula in the usual sense.

I think your posts will have to be moderated again.

Mark